What is wrong with this expression? "if (factorarray[x]%2 == 0 && factorarray[x]%3 ==0..."

Question

Here is the code:

#include <iostream>
#include <ctype.h>
#include <iomanip>


using namespace std;

class Project3
{
public:
    void factorcalc()
    {
        while (i<5000000){
            ++i;
            factor = prime/i;
            if (factor == (long long int)factor){
            //if (modf(primeoverthree, 0) == 0){
                cout << "Factor is: " << setprecision(12) << factor << endl;
                cout << "Denominator is: " << setprecision(12) << i << endl;

                ++x;
                factorarray[x] = factor;
            }
            else{/*empty else statement*/}
        }
    }
    void displayfactorresults ()
    {
        for (x=0; x<9; ++x)
            if (factorarray[x]%2 == 0 && factorarray[x]%3 == 0 && factorarray[x]%5 == 0 && factorarray[x]%7 == 0){
            cout << factorarray[x] << setprecision(12) << endl;
            }
    }
private:
    long double factor = 1;
    long double prime = 600851475143;
    long double i = 0;
    int x = 0;
    long double factorarray[9];
};


int main() {
    Project3 Pro3;
    Pro3.factorcalc();
    Pro3.displayfactorresults();

    return 0;
}

The code is in response to Project 3 on projecteuler.net. I am attempting to find the largest prime factor of 600851475143 using basic knowledge of c++ & common sense.

The error is occurring in the void displayfactorresults function.

The compiler outputs the error message:

"invalid operands to binary expression 'long-double' & 'long-double'"

Make your `factorarray` and other numbers be `uint64_t` . Using floating point for this task is hopeless — M.M, May 20 '15 at 01:53

Keith Thompson · Accepted Answer · 2015-05-20T02:08:25.887

The % remainder operator can only be applied to integer operands.

factorarray[x] is of type long double, a floating-point type.

Either use the fmod function (watch out for rounding errors!) or convert the operands to an integer type.

Or, better yet, make factorarray an array of some sufficiently large integer type in the first place. You're already assuming that the values can be converted to long long int. Converting something to a desired type is usually not as good as using the desired type in the first place.

score 1 · Answer 2 · answered May 20 '15 at 01:45

1

factorarray is an array of doubles. You cannot use modulo (%) on double. Either convert it to an array of int or cast the array element before you use it ((int)factorarray[x]) % 2 == 0

answered May 20 '15 at 01:45

Victor Hurdugaci

28,177
5
87
103

It may not be safe to assume that `factorarray[x]` can be converted to `int` without overflow. – Keith Thompson May 20 '15 at 01:46

score 1 · Answer 3 · answered May 20 '15 at 05:51

1

factorarray[x] % 2 == 0 (double type % 2 == 0) makes no sense because the reminder would be never 0, and of course compiler wont allow to do so.

answered May 20 '15 at 05:51

Matrix Buster

299
3
9

What is wrong with this expression? "if (factorarray[x]%2 == 0 && factorarray[x]%3 ==0..."

3 Answers3